For the reaction, `A(s) + 3B(g) rarr 4C(g) + d(l) Delta H and Delta U` are related as :

Options

A. `Delta H = DeltaU`
B. `Delta H = DeltaU + 3RT`
C. `Delta H = DeltaU + RT`
D. `Delta H = DeltaU - 3RT`

Explanation

To solve the problem regarding the relationship between ΔH (change in enthalpy) and ΔU (change in internal energy) for the reaction:

\[ A(s) + 3B(g) \rightarrow 4C(g) + D(l) \]

we can follow these steps:

### Step 1: Understand the Definitions
- **ΔH** represents the change in enthalpy of the reaction.
- **ΔU** represents the change in internal energy of the reaction.

### Step 2: Identify the Gaseous Reactants and Products
In the given reaction:
- Reactants: 3B(g) (gaseous)
- Products: 4C(g) (gaseous)
- Note: A(s) is solid and D(l) is liquid, so they do not contribute to ΔN_g.

### Step 3: Calculate ΔN_g
ΔN_g is defined as the change in the number of moles of gaseous products minus the change in the number of moles of gaseous reactants.

\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]

From the reaction:
- Moles of gaseous products = 4 (from 4C)
- Moles of gaseous reactants = 3 (from 3B)

Thus,
\[ \Delta N_g = 4 - 3 = 1 \]

### Step 4: Use the Relationship Between ΔH and ΔU
The relationship between ΔH and ΔU can be expressed as:

\[ \Delta H - \Delta U = \Delta N_g \cdot R \cdot T \]

Where:
- R is the universal gas constant (approximately 8.314 J/mol·K)
- T is the temperature in Kelvin

### Step 5: Substitute ΔN_g into the Equation
Substituting the value of ΔN_g into the equation, we have:

\[ \Delta H - \Delta U = 1 \cdot R \cdot T \]

This simplifies to:

\[ \Delta H - \Delta U = R \cdot T \]

### Step 6: Rearranging the Equation
Rearranging the equation to express ΔH in terms of ΔU gives:

\[ \Delta H = \Delta U + R \cdot T \]

### Conclusion
The relationship between ΔH and ΔU for the given reaction is:

\[ \Delta H = \Delta U + R \cdot T \]

### Final Answer
The correct option is that ΔH and ΔU are related as:

\[ \Delta H - \Delta U = R \cdot T \]